This problem is a Double Reactant. "K+ H20 ----> KOH + H2" would equal "2K + 2H20 ---> 2KOH + H2"
Balanced :2 K + 2 H2O ----> 2 KOH + H2
The balanced chemical equation for the reaction between potassium (K) and sulfuric acid (H2SO4) to form potassium sulfate (K2SO4) and hydrogen gas (H2) is: 2K + H2SO4 -> K2SO4 + H2
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and potassium (K) is: 2HCl + 2K -> 2KCl + H2.
The equilibrium constant (K eq) for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) would be [H2][Cl2]/[HCl]^2, where the square brackets indicate the molar concentrations of the respective species at equilibrium.
You need to have values for K, Br2 and 2KBr in order to find out the balance. For example, if you had the values as H2 + O2 = H20 the balanced chemical equation would be 10 H2 = H20.
k+h2o>koh+h2
Balanced :2 K + 2 H2O ----> 2 KOH + H2
The balanced chemical equation for the reaction between potassium (K) and sulfuric acid (H2SO4) to form potassium sulfate (K2SO4) and hydrogen gas (H2) is: 2K + H2SO4 -> K2SO4 + H2
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and potassium (K) is: 2HCl + 2K -> 2KCl + H2.
Balanced equation: 2 H2 + O2 --> 2 H2O
2 K + Br2 -> 2 Kbr
The equilibrium constant (K eq) for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) would be [H2][Cl2]/[HCl]^2, where the square brackets indicate the molar concentrations of the respective species at equilibrium.
You need to have values for K, Br2 and 2KBr in order to find out the balance. For example, if you had the values as H2 + O2 = H20 the balanced chemical equation would be 10 H2 = H20.
2K + 2H2O --> 2KOH + H2 The balanced equation shows that there is a 2 in front of water.
The balanced chemical equation for the reaction of mercury (Hg) with hydrochloric acid (HCl) is: Hg + 2HCl -> HgCl2 + H2
You need the balanced chemical reaction equation :------------------------------------------------------------------------2 K + 2 H2O ----. 2 KOH + H2moles H2 = n H2 = ( 1.60 L ) ( 1 mol H2 / 22.4 L H2 ) = 0.07143 moles H2moles K = n K = ( 0.07143 mol H2 ) ( 2 mol K / 1 mol H2 ) = 0.14286 moles Kmass K = m K = ( 0.14286 mol K ) ( 39.10 g K / mol K )mass K = m K = 5.56 g of potassium
Halving the concentration of H2 will decrease the rate of the reaction, assuming it is a first-order reaction with respect to H2. Since the rate law is rate = k[NO]^2[H2], cutting the concentration of H2 in half will decrease the rate of the reaction by a factor of 0.5.